Question

Two point charges +q and +4q are separated by a distance of 6a. find the point on the line joining the two charges where the electric field is zero

Answer 1

Answer:

2a

Explanation:

Two point charges +q and +4q are separated by a distance of 6a.

$\vec{E}=\dfrac{\vec{F}}{q}$

Let a charge Q place x distance from +q charge.

Force between +q and Q

$F=\dfrac{kqQ}{x^2}$

$E=\dfrac{kq}{x^2}$

Force between +4q and Q

$F=\dfrac{kqQ}{(6a-x)^2}$

$E=\dfrac{k\cdot4q}{(6a-x)^2}$

The point on the line joining the two charges where the electric field is zero.

Both filed must be same.

$\dfrac{kq}{x^2}=\dfrac{k\cdot4q}{(6a-x)^2}$

$\dfrac{1}{x^2}=\dfrac{2^2}{(6a-x)^2}$

$\dfrac{1}{x}=\dfrac{2}{6a-x}$

$6a-x=2x$

$x=2a$

Hence, The point is 2a distance from +q charge.

Answer 2

Answer:

Answer : 2a

Explanation:

Electric field is zero at P

if the field due to +q balance the field due to +4q

1 / 4π∈0 * q / x2 = 1 / 4π∈0 * 4q /( 6a - x) ^2

1 / x^2 = 4 / ( 6a - x) ^ 2

Taking square root on both sides

1 / x = 2 / 6a - x

On cross multiplying we get ,

2x = 6a - x

2x + x = 6a

3x = 6a

x = 2a

Hence the required point is at a distance of 2a from q

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